Difference between revisions of "Differential equation"

Revision as of 19:40, 6 March 2022

A differential equation is a functional equation involving functions and their derivatives.

The order of a differential equation is the largest order of any derivative that appears in the equation.

Examples

$f(x) = f'(x)$ has solutions $Ce^x$ for all real constants $C$ .

$f(x) = -f''(x)$ has solutions $C \cos(x + a)$ for all real constants $C$ and $a$ . The solutions with $a = 0$ are $C \cos x$ ; those with $a = \frac{\pi}{2}$ are $C \sin x$ .

Solutions

Separation of variables is a convenient technique for solving certain types of differential equations. Essentially, the method involves rewriting the equation so that each side is an expression in only one variable and then taking the antiderivative of both sides.

When solving differential equations, it is best to notate functions using a single variable name instead of spelling out the function and its arguments, for example using $y$ instead of $f(x)$ . Here we also use Leibniz notation $\frac{dy}{dx}$ for the derivative because it allows for manipulating $dy$ and $dx$ individually.

Worked example

To solve the differential equation $\frac{dy}{dx} + y = 2xy + x^2\frac{dy}{dx},$ we manipulate and factor to get $(1 - x^2)\frac{dy}{dx} = (2x - 1)y,$ then rearrange: $\frac{1}{y} \: dy = \frac{2x - 1}{1 - x^2} \: dx.$ We then proceed to take the antiderivatives $\int \frac{1}{y} \: dy = \int \frac{2x - 1}{1 - x^2} \: dx,$ at which point the equation can be solved using integration techniques.

Approximations

Euler's method uses repeated tangent-line approximations to approximate the solution to first-order differential equations.

Constant expressions

Certain expressions involving solutions to differential equations can be proven constant by noting that their derivatives are always $0$ . These constant expressions can then be used to prove properties of the solutions.

For example, when $f(x) = -f''(x)$ , $\begin{align*} \left( f(x)^2 + f'(x)^2 \right)' &= \left( f(x)^2 \right) ' + \left( f'(x)^2 \right) ' \\ &= 2f(x)f'(x) + 2f'(x)f''(x) \\ &= 2f'(x) \left( f(x) + f''(x) \right) \\ &= 0. \end{align*}$ Using $\sin' x = \cos x$ allows for reconstructing the familiar identity $\sin^2 x + \cos^2 x = \sin^2 0 + \cos^2 0 = 0^2 + 1^2 = 1$ for all real $x$ .

When $f(x) = f'(x)$ , for any real constant $S$ , $\begin{align*} \left( f(x)f(S-x) \right)' &= f'(x)f(S-x) + f(x)(f(S-x))' \\ &= f'(x)f(S-x) + f(x)(S-x)'f'(S-x) \\ &= f(x)f(S-x) + f(x)(-1)f(S-x) \\ &= 0. \end{align*}$ Letting $S = a + b$ and evaluating at both $x = 0$ and $x = a$ gives $f(a)f(b) = f(0)f(a+b),$ which using $e^0 = 1$ becomes the familiar identity $e^ae^b = e^{a+b}$ for all real $a$ and $b$ .

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 ===Worked example===
-To solve the differential equation <cmath>\frac{dy}{dx} + y = 2xy + x^2\frac{dy}{dx},</cmath> we manipulate and factor to get <cmath>(1 - x^2)\frac{dy}{dx} = (2x - 1)y,</cmath> then rearrange: <cmath>\frac{1}{y} \: dy = \frac{2x - 1}{1 - x^2} \: dx.</cmath> We then proceed to take the antiderivatives <cmath>\int \frac{1}{y} \: dy = \int \frac{2x - 1}{1 - x^2} \: dx,</cmath>
+To solve the differential equation <cmath>\frac{dy}{dx} + y = 2xy + x^2\frac{dy}{dx},</cmath> we manipulate and factor to get <cmath>(1 - x^2)\frac{dy}{dx} = (2x - 1)y,</cmath> then rearrange: <cmath>\frac{1}{y} \: dy = \frac{2x - 1}{1 - x^2} \: dx.</cmath> We then proceed to take the antiderivatives <cmath>\int \frac{1}{y} \: dy = \int \frac{2x - 1}{1 - x^2} \: dx,</cmath> at which point the equation can be solved using [[Integral|integration techniques]].
 ==Approximations==

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